the lenght of a species of fish is normally distributed with a mean of 24.3 inches and a standard deviation of 1.8 inches. What is the lenght that is exceeded by 30%of these fish

http://davidmlane.com/hyperstat/z_table.html

You will have to determine z-score (use a z-table with the information you are given), then use the z-score formula to find x (the length).

Z-score formula is this:

z = (x - mean)/sd
mean = 24.3
sd = 1.8

Once you find the z-score in the table, substitute the z-score and the data above into the formula and solve for x.

To find the length that is exceeded by 30% of these fish, we need to look for the length that corresponds to the 70th percentile of the distribution.

First, we need to standardize the desired percentile using the standard normal distribution (z-score).

To calculate the z-score, we can use the formula:

z = (X - μ) / σ

where:
X = desired length
μ = mean of the distribution (24.3 inches)
σ = standard deviation of the distribution (1.8 inches)

Now, we need to find the z-score corresponding to the 70th percentile. We can use a standard normal distribution table or calculators to determine this value. The z-score corresponding to the 70th percentile is approximately 0.524.

Now that we have the z-score, we can solve for the desired length (X) by rearranging the formula:

X = μ + (z * σ)

Plugging in the values:

X = 24.3 + (0.524 * 1.8)

X ≈ 24.3 + 0.9432

X ≈ 25.2432

Therefore, the length that is exceeded by 30% of these fish is approximately 25.24 inches.